How do we derive the expression for the potential due to a point charge at a distance r away from it (taking potential at infinity to be zero) ?

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integrate the work done in bringing a unit charge from infinity to that point at distance r using coulomb's law

thanks quakine, but this is where i'm struck I take Q due to which i need to calculate potential at point A I bring a test charge from infinity to the point A without any acceleration ! we have \[w = \int\limits_{\infty}^{A}F.dl\] Here F is the external force that an external system applies to bring the test charge to A let the test charge be a unit positive test charge. the force F is equal and opposite to that of the force due to Q therefore I substitute F= -E { E is the electric field due to Q } therefore we've \[w = -\int\limits_{\infty}^{A}E.dl\] now I substitute for E due to Q , i.e\[E=Q/4\Pi \in0 r ^{2} \] direction of E is opposite to the displacement dl ! here's i'm struck !

Just solve the integral and substitute the extreme values, taking into account that the value in infinite is zero. Some clue: the integral function of (1/r^2)dr is -1/r

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